Question

Use the squeeze theorem to prove that limit as n->infinity of (e/3)^n is convergent at 0

Answer 1

@daniel.ohearn1 hey if you don't mind, could you please tell me how you proved this? I just started learning this stuff so Im a bit confused lol

Answer 2

The squeeze theorem says that if two equations are bounded above and below this function and they both have have limits approaching in this case infinity then their limits are the same as it's limit.. So one can simply experiment with you calculator until you find two functions that sandwich at 0..

Answer 3

oh but don't we have to show it through manipulating limits?

Answer 4

cus its a prove question

Answer 5

Usually with Squeeze Theorem Questions u will have two functions whose Limits you can easily calculate algebraically or L'Hos and one obscure limit, then proving they all have the same Limit is more exciting. In this case I just opted to prove it by Squeeze Theorem because there's a statement given later in the book with sequences the Limit of the sequence r^n is 0 if -1<r<or=1 and L is 1 if r=1. The sequence is convergent...

Answer 6

oh okay but if this were an exam question, just out of curiosity, how would you structure your response?

Answer 7

I would judge that it is 0 by looking at the graph and by looking at other limits
lim (x)/2x x->inf ; lim (x^2/2x^2) x->inf ; U can notice that e is less then 3 so the denominator will approach infinity faster than the numerator...

Answer 8

ohh okayyy that makes sense, thanks!

Answer 9

Yep, your welcome